Problem: Let $g(x)=-4\log_3(x)$. Find $g'(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{4}{x\ln(3)}$ (Choice B) B $-\dfrac{4}{x\log_3(x)}$ (Choice C) C $-\dfrac{4}{x}$ (Choice D) D $-\dfrac{4\ln(x)}{\ln(3)}$
Answer: The expression for $g(x)$ includes a logarithmic term. Remember that the derivative of the general logarithmic term $\log_a(x)$ (where $a$ is any positive constant and $a\neq 1$ ) is $\dfrac{1}{\ln(a)\cdot x}$. Put another way, $\dfrac{d}{dx}[\log_a(x)]=\dfrac{1}{\ln(a)\cdot x}$. [Is there an easy way to memorize that?] We can use this to find the derivative of the function as shown below. $\begin{aligned} g'(x)&=\dfrac{d}{dx}[-4\log_3(x)] \\\\ &=-4\dfrac{d}{dx}[\log_3(x)] \\\\ &=-4\cdot\dfrac{1}{\ln(3)x} \\\\ &=-\dfrac{4}{x\ln(3)} \end{aligned}$ In conclusion, $g'(x)=-\dfrac{4}{x\ln(3)}$.